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I'm dealing with decomposition of the quotient ring ${{\mathbb{Z}_2}\left[ {x,y} \right]}$ over ${\left( {{x^3} - 1,{y^3} - 1} \right)}$. I know that I should use the Chinese remainder theorem but I failed. How can this ring be decomposed?

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  • $\begingroup$ While saying you tried this and failed is OK, it would be a lot more helpful to know what steps you took and how that failed. $\endgroup$ – rschwieb Nov 28 '13 at 23:55
  • $\begingroup$ Is it a requirement that you apply the CRT or are you just saying that's what you thought of? $\endgroup$ – rschwieb Nov 29 '13 at 13:20
  • $\begingroup$ It is not a requirement. Indeed, I want to determine the zero-divisors and units in the ring ${\mathbb{Z}{_p}\left[ {x,y} \right]}$ over ${\left\langle {{x^n} - 1,{y^m} - 1} \right\rangle }$ $\endgroup$ – egrtomath Nov 29 '13 at 14:32
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Here's a hint for the direction using the Chinese Remainder theorem:

Natural candidates for ideals would be $(x-1,y-1),(x-1,y^2+y+1),(x^2+x+1,y-1),(x^2+x+1,y^2+y+1)$.

This decomposes the ring into four subrings, three of which are fields. The reducible one splits into two fields. (Your job is to figure out all of these fields :) )

Another way to go about it is to look at the idempotents. This is particularly easy to compute in a ring with characteristic $2$ since $p(x,y)^2=p(x^2,y^2)$ for any polynomial here. After checking what an idempotent has to look like, you quickly find out there are 32 idempotents. Another way to approach this is as a group ring $F_2[C_3\times C_3]$. This helps with finding idempotents because in this case, the sum of elements in any subgroup of $C_3\times C_3$ is idempotent.

In a few minutes I whipped up a program to find the five primitive idempotents. Working by pencil and wits to compute this is still possible.

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